Show that Area form is independent of parameterization?

In summary, the question is how to show that the area form in differential geometry, √(EG-F2)du\wedgedv, is independent of the choice of local parameterization, where E, F, and G are coefficients of the first fundamental form. The poster is asking for ideas on how to approach this problem, and also wondering why this independence does not hold for standard integrals in ℝn. They are attempting to change from one parameterization to another and calculate the formula by definition to see if they get the same result. This involves using the determinants of Jacobian matrices, which eventually cancel out to give the same result.
  • #1
phyalan
22
0
In differential geometry, how can one show that the area form: √(EG-F2)du[itex]\wedge[/itex]dv is independent of the choice of local parameterization?
Here E,F,G are the coefficients of first fundamental form. Please someone gives me some ideas.
 
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  • #2
phyalan said:
In differential geometry, how can one show that the area form: √(EG-F2)du[itex]\wedge[/itex]dv is independent of the choice of local parameterization?
Here E,F,G are the coefficients of first fundamental form. Please someone gives me some ideas.

Maybe if you give us the definition of E, G, F and/or how they are calculated, that
would help.

And, BTW : why doesn't the independence of parametrization hold for standard
integrals in standard integration in ℝn , where the area is scaled by
the determinant of the Jacobian J(f) of the change of variables?
 
  • #3
Actually, what I am trying to do is to change from one parameterization to another and calculate the formula by definition to see if they give the same result under different parameterization, but I am not sure I am doing something valid. For instance, E=[itex]\phi_{u}\cdot\phi_{u}[/itex] where[itex]\phi[/itex] is a local parameterization from a open set in R^2 to the surface concerned and [itex]\phi_{u}=\partial \phi / \partial u\circ\phi^{-1}[/itex] so can I write [itex]\phi_{u}=\phi_{v}\partial v/\partial u [/itex]for another parameterization? In that case I can express the thing with determinants of Jacobian and they eventually cancel out to give same result.
 

Related to Show that Area form is independent of parameterization?

1. What is the area form in parameterization?

The area form in parameterization is a mathematical concept that represents the amount of area enclosed by a parameterized curve or surface. It is typically denoted by dA or dS for curves and surfaces respectively.

2. How is the area form independent of parameterization?

The area form is independent of parameterization because it is a geometric property that is not affected by how the curve or surface is parametrized. This means that as long as the curve or surface remains the same, the value of the area form will not change.

3. Why is it important to show that the area form is independent of parameterization?

It is important to show that the area form is independent of parameterization because it allows us to use different parametrizations for the same curve or surface without affecting the calculations of its area. This makes it a useful tool in various mathematical and scientific fields.

4. What is the proof that the area form is independent of parameterization?

The proof that the area form is independent of parameterization involves using the Jacobian determinant of the parametrization to show that it cancels out when calculating the area form. This demonstrates that the value of the area form remains the same regardless of the parametrization used.

5. How does the independence of the area form relate to the concept of orientation?

The independence of the area form is closely related to the concept of orientation. In a mathematical context, orientation refers to the direction in which a curve or surface is traversed. The independence of the area form allows us to calculate the same area regardless of the orientation of the curve or surface, as long as the parametrization remains the same.

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